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Let $\ A,B $ be a poisson process with $\ \lambda = 2 $ in a timeframe of $\ 1 $ minute. Both are independent variables. Let $\ T $ = time that has passed from $\ 0 $ until the occurrence of the first event of $\ A $

I need to compute $\ P(T > 0.25) $

According to the solution given, $\ T \sim exp(2) $ and I can not understand why.

I found those two questions : https://stats.stackexchange.com/questions/2092/relationship-between-poisson-and-exponential-distribution & What is the difference between a Poisson and an Exponential distribution?

So I understand that time between two events of a poisson process is exponentially distributed yet I can't understand how to compute the parameter? In this question it means I have a event every 30 seconds. So if I understand correctly the time between events has a mean of $\ 30 $ seconds so $\ T \sim exp(30) $ ?

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Note $T>t$ if and only if there are no occurrences in the interval $[0,t]$. The number of occurrences in $[0,t]$ has a $\text{Poi}(2t)$ distribution. Therefore, $$ P(T>t)=P(\text{Poi}(2t)=0)=e^{-{2t}}\quad\implies\quad P(T\le t)=1-e^{-2t} $$ Note that this is exactly the cdf of an $\text{Exp}(2)$ distribution.

Here is why $T$ is not distributed like $\text{Exp}(30\text{ sec})$. The parameter $\lambda$ is not the mean of $T$; the mean of $T$ is $1/\lambda$. Since, as you intuited, $T$ has a mean of $30\text{ sec}=1/2\text{ min}$, it follows, that $E[T]=1/\lambda=1/2$, so $\lambda=2$. Think of $\lambda$ as a "rate" parameter. $T$ is a wait time, and the larger $\lambda$ is, the faster that this time passes.

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